#13 ### Educational Material: Number Theory and Primitive Roots#### Problems and Exercises**9.** Prove that if \( n \) is a perfect square, then \( \lambda(n) = 1 \).**10.** Let \( r \) be a primitive root of some \( n \geq 3 \). Prove that:\[r^{\frac{\phi(n)}{2}} \equiv \pm 1 \pmod{n}\]**11.** Find \( \tau(n), \sigma(n), \lambda(n), \mu(n), \omega(n), \) and \( \phi(n) \) for the following integers:- 2250- 199- 286936650- 22!**12.** Let \( p = 17 \) and \( d \) be a divisor of \( \phi(p) \). Determine \( \psi(d) \) for each \( d \). List all divisors, \( d \), of \( \phi(p) \).**13.** Calculate all the primitive roots of 41 and 26.**14.** Demonstrate that 21 has no primitive root. - Notes show attempts considering primes such as 2, 3, 5, 7, 11, 21.**15.** Let \( r \) be a primitive root of \( n \). If \(\gcd(a, n) = 1\), then the smallest power \( k \) such that \( r^k \equiv a \pmod{n} \) is called the **index of \( a \) relative to \( r \)**, denoted by \(\text{ind}_r(a)\). This concept is useful to solve congruences. Consider the properties of indices.- Solve: \( 8x^4 \equiv 11 \pmod{13} \).#### Annotations and Remarks- The problems are sequentially numbered with some highlighted or circled for emphasis.- There are handwritten notes alongside problems indicating checks, attempted solutions, or additional thoughts like "150, 157 maybe theorem".

(a) Consider a=41. Therefore: φa=φ41=411-410=40 Now φa=40 can be written as: φa=40=23×5. Therefore the powers for tests are s=402=20 and s=405=8.We know that n is a primitive root, if n20≠1mod41 and n8≠1mod41. Now consider primitive tests: Let n=2: 220=1mod41, Hence n=2 is not a primitive root. Let n=3: 320=40mod41 and 38=1mod41 Hence n=3 is not a primitive root. Let n=4: 420=1mod41, Hence n=4 is not a primitive root. Let n=5: 520=1mod41, Hence n=5 is not a primitive root. Let n=6: 620=40mod41 and 68=10mod41. Hence n=6 is a primitive root.Therefore n=6 is a least primitive root of 41. If we check similarly all numbers up to n=40, we will get all primitive roots of 41, which are listed below. All the primitive root of 41 are: 6, 7, 11, 12, 13, 15, 17, 19, 22, 24, 26, 28, 29, 30, 34, and 35.(b) Consider a=26. Therefore: φa=φ26=131-13021-20=12 Now φa=12 can be written as: φa=12=22×3. Therefore the powers for tests are s=122=6 and s=123=4.We know that n is a primitive root, if n6≠1mod26 and n4≠1mod12. Now consider primitive tests: Let n=2: 26=12mod26 and 24=16mod26. Hence n=2 is a primitive root of 26. Let n=3: 36=1mod26 Hence n=3 is a not primitive root of 26. Similarly the following are primitive root of 26: 7,11, 15, and , 19.